= Solution
The <modular inverse> of $2$ modulo $3$ is $2$, so the second <modular congruence> is $x\equiv2\pmod3$. The third means $10\mid2(x-2)$, equivalently $5\mid x-2$. Cancelling this common factor changes the modulus; it gives $x\equiv2\pmod5$.
The first three conditions consequently combine to $x\equiv2\pmod{15}$ and $x$ odd, hence $x=17+30k$ for an <integer> $k$. The last condition becomes
$$
30k\equiv10-17=-7\pmod{67}.
$$
As $30\cdot38=17\cdot67+1$, multiplying by the <modular inverse> $38$ gives $k\equiv2\pmod{67}$. Thus
$$
\boxed{x=77+2010\ell,\qquad\ell\in\mathbb Z.}
$$
The <Chinese remainder theorem> states that <modular congruences> with pairwise coprime moduli have a unique solution modulo their product. Here those moduli are $2,3,5,67$, with product $2010$. The derivation proves necessity, and substitution gives $77\equiv1\pmod2$, $2\cdot77\equiv1\pmod3$, $2\cdot77\equiv4\pmod{10}$ and $77\equiv10\pmod{67}$, proving sufficiency for the whole displayed class.
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